Integration using Partial Fractions MCQ Quiz - Objective Question with Answer for Integration using Partial Fractions - Download Free PDF

Calculation:

We know that the integral is of the form:

\( f(x) = \frac{1}{2} \log_e \left( \frac{x^2 + 3}{x^2 + 1} \right) + C \)

Now, we substitute the given value of f(3) to find the constant C :

⇒ \( f(3) = \frac{1}{2} \log_e \left( \frac{6}{5} \right) + C \)

We are given that:

⇒ \( f(3) = \frac{1}{2} (\log_e 5 - \log_e 6) \)

Equating the two expressions for f(3):

⇒ \( \frac{1}{2} \log_e \left( \frac{6}{5} \right) + C = \frac{1}{2} (\log_e 5 - \log_e 6) \)

Since both sides are equal, we conclude that C = 0 .

Thus, the function becomes:

⇒ \( f(x) = \frac{1}{2} \log_e \left( \frac{x^2 + 3}{x^2 + 1} \right) \)

Now, we can calculate f(4):

⇒ \( f(4) = \frac{1}{2} \log_e \left( \frac{16 + 3}{16 + 1} \right) = \frac{1}{2} \log_e \left( \frac{19}{17} \right) \)

Thus, the value of f(4) is:

⇒ \( \frac{1}{2} \log_e \left( \frac{19}{17} \right) \)

 \( \frac{1}{2} (\log_e 19 - \log_e 17) \)

Hence, the correct answer is Option 1.

Given:

\(\rm \int\frac{2x+3}{x^3+x^2-2x}dx\)

Concept:

Use concept of partial fractions

\(\frac{f(x)}{g(x)\cdot h(x)}=\frac{A}{g(x)}+\frac{B}{h(x)}\)

And use formula of integration

\(\rm \int\frac{1}{x}\ dx=log|x|+c\)

Calculation:

\(\rm \int\frac{2x+3}{x^3+x^2-2x}dx\)

\(\rm =\int\frac{2x+3}{x(x-1)(x+2)}dx\)

Use concept of partial fractions

\(\rm \frac{2x+3}{x(x-1)(x+2)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+2}\)

Now, Cross multiply by denominators 

\(\rm 2x+3=A(x^2+x-2)+B(x^2+2x)+C(x^2-x)\)

Compare the coefficients on both the sides.

\(\rm A+B+C=0\) .........(1)

\(\rm A+2B-C=2\) ...........(2) and

\(\rm -2A=3\)

\(\rm \implies A=\frac{-3}{2}\)

On adding equation (1) and (2)

\(\rm 2A+3B=2\)

Put value of A then

\(\rm 2\times\frac{-3}{2}+3B=2\)

\(\rm \implies B=\frac{5}{3}\)

Now put the value of A and B in equation (1) then we get

\(\rm C=\frac{-1}{6}\)

Now put all these values in integral then 

\(\rm \int\frac{2x+3}{x^3+x^2-2x}dx\)

\(\rm =\int\frac{-3}{2}.\frac{1}{x}\ dx+\int\frac{5}{3}.\frac{1}{x-1}\ dx+\int\frac{-1}{6}.\frac{1}{x+2}\ dx\)

\(\rm =-\frac{3}{2}log|x|+\frac{5}{3}log|x-1|-\frac{1}{6}log|x+2|+c\)

\(\rm =\frac{5}{3}log|x-1|-\frac{3}{2}log|x|-\frac{1}{6}log|x+2|+c\)

Hence the option (2) is correct.

Calculation:

Given:

\(\int \frac{x}{(x-1)(x-2)} dx\)

\(\frac{x}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}\)

⇒ \(x = A(x-2) + B(x-1)\)

\(1 = A(1-2) + B(1-1) \Rightarrow A = -1\)

\(2 = A(2-2) + B(2-1) \Rightarrow B = 2\)

The integral becomes:

⇒ \(\int \left( \frac{-1}{x-1} + \frac{2}{x-2} \right) dx\)

⇒ \(-\int \frac{1}{x-1} dx + 2\int \frac{1}{x-2} dx\)

⇒ \( -\ln|x-1| + 2\ln|x-2| + C\)

⇒ \( \ln\left| \frac{(x-2)^2}{x-1} \right| + C\)

Hence option 2 is correct

Calculation

Given:

Let \(f(x) = \int \frac{x}{(x^2+1)(x^2+3)} dx\)

Using partial fraction decomposition:

\(\frac{x}{(x^2+1)(x^2+3)} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+3}\)

Since the numerator is just 'x', B and D must be zero.

\(\frac{x}{(x^2+1)(x^2+3)} = \frac{Ax}{x^2+1} + \frac{Cx}{x^2+3}\)

Multiplying both sides by \((x^2+1)(x^2+3)\):

\(x = Ax(x^2+3) + Cx(x^2+1)\)

\(x = Ax^3+3Ax + Cx^3+Cx\)

\(x = (A+C)x^3 + (3A+C)x\)

Comparing coefficients:

\(A+C = 0 \implies C = -A\)

\(3A+C = 1\)

Substituting \(C=-A\):

\(3A-A = 1\)

\(2A = 1 \implies A = \frac{1}{2}\)

\(C = -\frac{1}{2}\)

\(f(x) = \int \left(\frac{1/2 x}{x^2+1} - \frac{1/2 x}{x^2+3}\right) dx\)

\(f(x) = \frac{1}{2} \int \frac{x}{x^2+1} dx - \frac{1}{2} \int \frac{x}{x^2+3} dx\)

Let \(u = x^2+1 \implies du = 2x dx\)

Let \(v = x^2+3 \implies dv = 2x dx\)

\(f(x) = \frac{1}{4} \int \frac{du}{u} - \frac{1}{4} \int \frac{dv}{v}\)

\(f(x) = \frac{1}{4} \ln|u| - \frac{1}{4} \ln|v| + K\)

\(f(x) = \frac{1}{4} \ln|x^2+1| - \frac{1}{4} \ln|x^2+3| + K\)

\(f(x) = \frac{1}{4} \ln\left|\frac{x^2+1}{x^2+3}\right| + K\)

Given \(f(3) = \frac{1}{4} \log\left(\frac{5}{6}\right)\)

\(f(3) = \frac{1}{4} \ln\left|\frac{3^2+1}{3^2+3}\right| + K\)

\(f(3) = \frac{1}{4} \ln\left|\frac{10}{12}\right| + K\)

\(f(3) = \frac{1}{4} \ln\left|\frac{5}{6}\right| + K\)

\(\frac{1}{4} \log\left(\frac{5}{6}\right) = \frac{1}{4} \log\left(\frac{5}{6}\right) + K \implies K=0\)

\(f(x) = \frac{1}{4} \ln\left|\frac{x^2+1}{x^2+3}\right|\)

Find f(0):

\(f(0) = \frac{1}{4} \ln\left|\frac{0^2+1}{0^2+3}\right|\)

\(f(0) = \frac{1}{4} \ln\left|\frac{1}{3}\right|\)

∴ \(f(0) = \frac{1}{4} \log\left(\frac{1}{3}\right)\)

Hence option 1 is correct

Calculation

Given the integral:

\(\int \frac{\log(1+x^4)}{x^3} \, dx = f(x) \log \left( \frac{1}{g(x)} \right) + \tan^{-1}(h(x)) + c\)

Let \(u = \log(1+x^4)\) and \(dv = \frac{1}{x^3} dx\).

Then \(du = \frac{4x^3}{1+x^4} dx\) and \(v = -\frac{1}{2x^2}\).

Using integration by parts:

\(\int \frac{\log(1+x^4)}{x^3} dx = uv - \int v du\)

\(= -\frac{\log(1+x^4)}{2x^2} - \int \left(-\frac{1}{2x^2}\right) \frac{4x^3}{1+x^4} dx\)

\(= -\frac{\log(1+x^4)}{2x^2} + \int \frac{2x}{1+x^4} dx\)

Let \(x^2 = t\), so \(2x dx = dt\).

The integral becomes:

\(\int \frac{2x}{1+x^4} dx = \int \frac{dt}{1+t^2} = \tan^{-1}(t) + C = \tan^{-1}(x^2) + C\)

\(\int \frac{\log(1+x^4)}{x^3} dx = -\frac{\log(1+x^4)}{2x^2} + \tan^{-1}(x^2) + C\)

Comparing with \(f(x) \log \left( \frac{1}{g(x)} \right) + \tan^{-1}(h(x)) + c\):

\(f(x) = \frac{1}{2x^2}\)

\(g(x) = 1+x^4\)

\(h(x) = x^2\)

\(f\left(\frac{1}{x}\right) = \frac{1}{2\left(\frac{1}{x}\right)^2} = \frac{x^2}{2}\)

\(f(x) + f\left(\frac{1}{x}\right) = \frac{1}{2x^2} + \frac{x^2}{2} = \frac{1+x^4}{2x^2}\)

\(h(x) \left[ f(x) + f\left( \frac{1}{x} \right) \right] = x^2 \left( \frac{1+x^4}{2x^2} \right) = \frac{1+x^4}{2}\)

∴ \(h(x) \left[ f(x) + f\left( \frac{1}{x} \right) \right] = \frac{1+x^4}{2} = \frac{g(x)}{2}\)

Concept:

Partial Fraction:

Calculation:

Here we have to find the value of \(\smallint \frac{{dx}}{{x\;\left( {x + 2} \right)}}\)

Let \(\frac{1}{{x\;\left( {x + 2} \right)}} = \frac{A}{x} + \frac{B}{{x + 2}}\)

⇒ 1 = A (x + 2) + B x --------(1)

By putting x = 0 on both the sides of (1) we get A = 1/2

By putting x = - 2 on both the sides of (1) we get B = - 1/2

\(\Rightarrow \frac{1}{{x\;\left( {x + 2} \right)}} = \frac{1}{2x} - \frac{1}{{2x + 4}}\)

\(\Rightarrow \smallint \frac{{dx}}{{x\;\left( {x + 2} \right)}} = \frac{1}{2}\smallint \frac{{dx}}{x} - \frac{1}{2}\;\smallint \frac{{dx}}{{x + 2}}\;\)

As we know that \(\smallint \frac{{dx}}{x} = \log \left| x \right|\; + C\)  where C is a constant

\(\Rightarrow \smallint \frac{{dx}}{{x\;\left( {x + 2} \right)}} = \frac{1}{2}\log \left| {\frac{x}{{x + 2}}} \right| + C\;\)where C is a constant

Formula used:

\(\rm \int \frac{1}{x} dx = log \: x + C\)

log_ax - log_ay = \(\rm log_{a}\frac{x}{y}\)

Calculation:

\(\int \frac{dx}{x(x^2 + 1)}\)

⇒ \(\rm \int \frac{dx}{x} - \frac{xdx}{x^{2} + 1}\)

⇒ \(\rm \int \frac{dx}{x} - \frac{1}{2}\int \frac{2xdx}{x^{2} + 1}\)

⇒ ln x - \(\rm \frac{1}{2} ln |1 + x^{2}|\) + c

⇒ \(\frac{1}{2} \ln x^{2} -\)\(\rm \frac{1}{2} ln |1 + x^{2}|\) + c

⇒ \(\rm \frac{1}{2}ln\left(\frac{x^2}{x^2 + 1}\right) + C\)

∴ \(\int \frac{dx}{x(x^2 + 1)}\) is equal to 

\(\rm \frac{1}{2}ln\left(\frac{x^2}{x^2 + 1}\right) + C\).

Concept:

Using partial fraction method

\(\rm 1\over (x-a).(x -b) \) = \(A\over(x-a)\) + \(B\over(x - b)\)

\(\rm \int \frac{{dx}}{x} = \log \left| x \right|\; + c\)

Calculation:

I = \(\int \rm \frac{{dx}}{{(x-2)\;\left( {x - 1} \right)}}\)

⇒ \(\rm 1\over (x-2).(x -1) \) = \(A\over(x-2)\) + \(B\over(x - 1)\)

⇒ 1 = A(x - 1) + B(x - 2)

Compair cofficient both sides

Cofficient of x is  A + B = 0

Coffiecient of constant 1 = -A - 2B

Solving the equation we get 

A = 1, B = -1

⇒ \(\int \rm \frac{{dx}}{{(x-2)\;\left( {x - 1} \right)}}\) = \(\int {dx\over (x-2)} \) - \(\int {dx\over (x-1)} \)

⇒ log|x - 2| - log|x - 1| + c       

= \(\rm log \left|{(x-2)\over (x - 1)}\right|+c\)                             [∵ log m - log n = log(\(\rm \frac mn\))]

Concept:

Integral property:

• ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
• \(\rm∫ {1\over x} dx = \ln x\) + C
• ∫ ex dx = ex+ C
• ∫ ax dx = (ax/ln a) + C ; a > 0,  a ≠ 1
• ∫ sin x dx = - cos x + C
• ∫ cos x dx = sin x + C

Calculation:

I = \(\rm \int {5x\over x^2+3x-4}dx\)

I = \(\rm \int {5x\over (x-1)(x+4)}dx\)

I = \(\rm \int {(x+4)+(4x-4)\over (x-1)(x+4)}dx\)

I = \(\rm \int{1\over x-1}dx+\int{4\over x+4} dx\)

I = ln (x - 1) + 4 ln (x + 4) + c

Concept:

• \(\rm \int\frac{1}{x+a}dx=log|x+a|+C\)
• \(\rm \int\frac{1}{(x+a)^n}dx=\frac{1}{(1-n)(x+a)^{n+1}}+C\)

Calculation:​

To solve: \(\rm \int \frac{2x^3}{(x^2+7)^2}dx\) 

Let us put x² = t ⇒2xdx = dt in \(\rm \int \frac{2x^3}{(x^2+7)^2}dx\)  

⇒ \(\rm \int \frac{2x^3}{(x^2+7)^2}dx = \rm \int \frac{t}{(t+7)^2}dt\)

This integrand is a proper rational fraction. therefore, by using the form of partial fraction, we write

⇒ \(\rm \frac{t}{(t+7)^2}=\frac{A}{t+7}+\frac{B}{(t+7)^2}\)

⇒ t = At + 7A + B 

By comparing coefficient of t and constant terms on both sides, we get A = 1 and 7A + B = 0

By solving these equation, we get  A = 1 and B = -7

\(⇒ \rm \frac{t}{(t+7)^2}=\frac{1}{t+7}-\frac{7}{(t+7)^2}\)

⇒ \(\rm \int\frac{t}{(t+7)^2}dt=\int\frac{1}{t+7}dt-\int\frac{7}{(t+7)^2}dt\)

⇒ \(\rm \int\frac{t}{(t+7)^2}dt=log\left | t+7 \right |+\frac{7}{(t+7)}+C\) 

Now put t = x² in the above equation we get

⇒ \(\rm \int\frac{2x^3}{(x^2+7)^2}dx=log\left | x^2+7 \right |+\frac{7}{(x^2+7)}+C\)

Concept Used:
\( \int_a^b f(x) d x=\int_a^b f(a+b-x) d x \)
also, sin (π/2 - x) = cos x 
and cos (π/2 - x) = sin x 
\( \sin x=\frac{2 \tan x/2 }{1+\tan ^2 x / 2}, \) \(\cos x=\frac{1-\tan ^2 x / 2}{1+\tan ^2 x / 2} \)

Also, \(\int \frac{d x}{a^2-x^2}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+c\)

Calculation:

Let \(I=\int_0^{π / 2} \frac{\cos ^2 x}{\sin x+\cos x} d x \ldots(1) \)
 ⇒ \(I=\int_0^{π / 2} \frac{\cos ^2(π / 2-x)}{\sin \left(π_2-x\right)+\cos \left(π_2-x\right)} d x \)
 ⇒ \(I=\int_0^{π / 2} \frac{\sin ^2 x}{\sin x+\cos x} d x \cdots \text {(2)} \)
Adding (1) and (2)
 ⇒ \(2 I=\int_0^{π / 2} \frac{\sin ^2 x+\cos ^2 x}{\sin x+\cos x} d x \)
 ⇒ \(2 I=\int_0^{π / 2} \frac{1}{\sin x+\cos x} d x \)

Now, let \(I_1=\int \frac{1}{\sin x+\cos x} d x\)

\(⇒ I_1=\int \frac{d x}{\frac{2 \tan x / 2}{1+\tan ^2 x / 2}+\frac{1-\tan^2 x / 2}{1+\tan ^2 x / 2}} \)
⇒ \(I_1=\int \frac{1+\tan ^2 x / 2}{2 \tan x / 2+1-\tan ^2 x / 2} d x \)

⇒ \(I_1 =\int \frac{\sec ^2 x / 2 d x}{2 \tan x / 2+1-\tan ^2 x / 2} \)

Put tan x/2 = t
 ⇒ \(\frac{1}{2} \sec ^2 x / 2=\frac{d t}{d x} \)
 ⇒ \(\sec ^2 x/2 dx = 2dt\)

⇒ I_{1\( =2 \int \frac{d t}{2 t+1-t^2}\)}
⇒ I1 \(=-2 \int \frac{d t}{t^2-2 t-1}\)
⇒ I1\( =-2 \int \frac{dt}{t^2-2 t-1+1-1}\)
⇒ I1 \(=-2 \int \frac{dt}{(t-1)^2-(\sqrt{2})^2}\)
⇒ I1 \( =2 \int \frac{dt}{(\sqrt{2})^2-(t-1)^2}\)
 ⇒I1 \(=2 \frac{1}{2 \sqrt{2}} \log \left|\frac{\sqrt{2}+t-1}{\sqrt{2}-t+1}\right|+c \)
⇒ I1 \( =\frac{1}{\sqrt{2}} \log \left|\frac{\sqrt{2}+\tan x / 2-1}{\sqrt{2}-\tan x / 2+1}\right|+c\)
Using I₁, in I

\(\Rightarrow I=\frac{1}{2 \sqrt{2}}\left[\ln \left|\frac{\sqrt{2}+\tan \pi / 4-1}{\sqrt{2}-\tan \pi / 4+1}\right|-\ln \left|\frac{\sqrt{2}+\tan 0-1}{\sqrt{2}-\tan 0+1}\right|\right]\)
\(\Rightarrow I=\frac{1}{2 \sqrt{2}}\left[\ln \left|\frac{\sqrt{2}}{\sqrt{2}}\right|-\ln \left|\frac{\sqrt{2}-1}{\sqrt{2}+1}\right|\right]\)
\(\Rightarrow I=\frac{1}{2 \sqrt{2}}\left[\ln \left|\frac{(\sqrt{2}+1)^2}{(\sqrt{2})^2-1^2} \right|\right]\)
\(\Rightarrow I=\frac{1}{2 \sqrt{2}} \ln \left|\frac{2+1+2 \sqrt{2}}{2-1}\right|\)\(\Rightarrow I=\frac{1}{2 \sqrt{2}} \ln |3+2 \sqrt{2}|\)

Concept:

Partial Fraction:

Calculation:

Here we have to find the value of \(\smallint \frac{{dx}}{{{e^x} - 1}}\)

Let e^x = t and by differentiating ex = t with respect to x we get

⇒ e^x dx = dt or dx = dt/e^x = dt/t

\(⇒ \smallint \frac{{dx}}{{{e^x} - 1}} = \;\smallint \frac{{dt}}{{t\left( {t - 1} \right)}}\)

Let \(\frac{1}{{t\left( {t - 1} \right)}} = \frac{A}{t} + \frac{B}{{t - 1}}\)

⇒ 1 = A (t - 1) + B t ---------(1)

By putting t = 0 on both the sides of (1) we get A = - 1

By putting t = 1 on both the sides of (1) we get B = 1

\(\Rightarrow \frac{1}{{t\left( {t - 1} \right)}} = \frac{{ - 1}}{t} + \frac{1}{{t - 1}}\)

\(\Rightarrow \smallint \frac{{dt}}{{t\left( {t - 1} \right)}} = \; - \;\smallint \frac{{dt}}{t} + \;\smallint \frac{{dt}}{{t - 1}}\)

As we know that \(\smallint \frac{{dx}}{x} = \log \left| x \right|\; + C\)  where C is a constant

 \(\Rightarrow \smallint \frac{{dt}}{{t\left( {t - 1} \right)}} = \; - \log \left| t \right| + \log \left| {t - 1} \right| + C\)

\(= \log \left| {\frac{{t - 1}}{t}} \right| + C\)

By substituting  ex = t in the above equation we get

\(\Rightarrow \smallint \frac{{dx}}{{{e^x} - 1}} = \log \left| {\frac{{{e^x} - 1}}{{{e^x}}}} \right| + C\)

Concept:

∫ 1 dx = x + constant

\(\int\frac{1}{x}dx=log\ x+constant\)

Calculation:

Given:

Let, \(I=∫ \frac {x^2}{x^2 - 3x + 2}dx\)

\(I= ∫ (1+\frac{3x-2}{x^2-3x+2})dx\)

\(I= ∫ 1dx+∫\frac{3x+2}{x^2-3x-2}dx\).....(i)

\(\frac{3x-2}{x^2-3x+2}=\frac{3x-2}{(x-2)(x-1)}=\frac{A}{(x-2)}+\frac{B}{(x-1)}\) ....(ii)

(3x - 2) = A (x - 1) + B (x - 2)

for x = 1

(3 (1) - 2) = B (1 - 2)

B = -1

for x = 2

(3 (2) - 2) = A (2 - 1)

A = 4

from equation (ii)

\(\frac{3x-2}{x^2-3x+2}=\frac{4}{(x-2)}-\frac{1}{(x-1)}\)

Now from equation (i)

\(I= ∫ 1dx+∫\frac{4}{(x-2)}dx-∫\frac{1}{(x-1)}dx\)

x - log |x - 1| + 4 log |x - 2| + c

where c is an arbitrary constant.

Concept:

Partial Fraction:

Calculation:

Here we have to find the value of \(\smallint \frac{{dx}}{{x\left\{ {6 \cdot {{\left( {\log x} \right)}^2} + 7\log x + 2} \right\}}}\)

Let log x  = t and dx/x = dt

\(⇒ \smallint \frac{{dx}}{{x\left\{ {6 \cdot {{\left( {\log x} \right)}^2} + 7\log x + 2} \right\}}} = \;\smallint \frac{{dt}}{{\left( {6{t^2} + 7t + 2} \right)}}\)

\(⇒ \smallint \frac{{dt}}{{\left( {6{t^2} + 7t + 2} \right)}} = \;\smallint \frac{{dt}}{{\left( {2t + 1} \right)\left( {3t + 2} \right)}}\)

\(⇒ \frac{1}{{\left( {2t + 1} \right)\left( {3t + 2} \right)}} = \frac{A}{{2t + 1}} + \frac{B}{{3t + 2}}\)

⇒ 1 = A (3t + 2) + B (2t + 1) --------(1)

By putting t = - 1/2 on both the sides of (1) we get A = 2

By putting t = - 2/3 on both the sides of (1) we get B = - 3

\(\Rightarrow \frac{1}{{\left( {2t + 1} \right)\left( {3t + 2} \right)}} = \frac{2}{{2t + 1}} - \frac{3}{{3t + 2}}\)

\(\;\Rightarrow \smallint \frac{{dt}}{{\left( {2t + 1} \right)\left( {3t + 2} \right)}} = \;\smallint \frac{{2}}{{2t + 1}}dt - \;\smallint \frac{{3dt}}{{3t + 2}}\)

As we know  that \(\smallint \frac{{dx}}{x} = \log \left| x \right|\; + C\)  where C is a constant

\(\Rightarrow \;\smallint \frac{{dt}}{{\left( {2t + 1} \right)\left( {3t + 2} \right)}} = \log \left| {2t + 1} \right| - \log \left| {3t + 2} \right| + C\)

\(= \log \left| {\frac{{2t + 1}}{{3t + 2}}} \right| + C\)

By substituting log x  = t in the above equation we get

\(\Rightarrow \smallint \frac{{dx}}{{x\left\{ {6 \cdot {{\left( {\log x} \right)}^2} + 7\log x + 2} \right\}}} = \;\log \left| {\frac{{2\log x + 1}}{{3\log x + 2}}} \right| + C\)

Given:

\(\rm \int\frac{2x+3}{x^3+x^2-2x}dx\)

Concept:

Use concept of partial fractions

\(\frac{f(x)}{g(x)\cdot h(x)}=\frac{A}{g(x)}+\frac{B}{h(x)}\)

And use formula of integration

\(\rm \int\frac{1}{x}\ dx=log|x|+c\)

Calculation:

\(\rm \int\frac{2x+3}{x^3+x^2-2x}dx\)

\(\rm =\int\frac{2x+3}{x(x-1)(x+2)}dx\)

Use concept of partial fractions

\(\rm \frac{2x+3}{x(x-1)(x+2)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+2}\)

Now, Cross multiply by denominators 

\(\rm 2x+3=A(x^2+x-2)+B(x^2+2x)+C(x^2-x)\)

Compare the coefficients on both the sides.

\(\rm A+B+C=0\) .........(1)

\(\rm A+2B-C=2\) ...........(2) and

\(\rm -2A=3\)

\(\rm \implies A=\frac{-3}{2}\)

On adding equation (1) and (2)

\(\rm 2A+3B=2\)

Put value of A then

\(\rm 2\times\frac{-3}{2}+3B=2\)

\(\rm \implies B=\frac{5}{3}\)

Now put the value of A and B in equation (1) then we get

\(\rm C=\frac{-1}{6}\)

Now put all these values in integral then 

\(\rm \int\frac{2x+3}{x^3+x^2-2x}dx\)

\(\rm =\int\frac{-3}{2}.\frac{1}{x}\ dx+\int\frac{5}{3}.\frac{1}{x-1}\ dx+\int\frac{-1}{6}.\frac{1}{x+2}\ dx\)

\(\rm =-\frac{3}{2}log|x|+\frac{5}{3}log|x-1|-\frac{1}{6}log|x+2|+c\)

\(\rm =\frac{5}{3}log|x-1|-\frac{3}{2}log|x|-\frac{1}{6}log|x+2|+c\)

Hence the option (2) is correct.